# Octiles Calculator for ungrouped data with examples

## Octiles for ungrouped data

Octiles are the values of arranged data which divide whole data into eight equal parts. They are 7 in numbers namely $O_1,O_2, \cdots, O_7$. Here $O_1$ is first octile, $O_2$ is second octile, $O_3$ is third octile and so on.

The formula for $i^{th}$ octile is

$O_i =$ Value of $\bigg(\dfrac{i(n+1)}{8}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 7$

where,

• $n$ is the total number of observations.

## Octiles Calculator for ungrouped data

Use this calculator to find the octiles for ungrouped (raw) data.

Octile Calculator
Enter the X Values (Separated by comma,)
Which Octile? (Between 1 to 7)
Results
Number of Obs. (n):
Ascending order of X values :
Required Octile : O{{index}}

## How to calculate octiles for ungrouped data?

Step 1 - Enter the $x$ values separated by commas

Step 2 - Enter the nuber between 1 to 7 (inclusive)

Step 3 - Click on "Calculate" button to get octile for ungrouped data

Step 4 - Gives the output as number of observations $n$

Step 5 - Gives the output as ascending order data

Step 6 - Gives the required octile

## Octiles for ungrouped data Example 1

Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

65,76,64,73,74,80, 71, 68,66, 81, 79, 75, 70, 62, 83,63, 77, 78.

Find the value of $O_4$ and $O_6$.

#### Solution

The formula for $i^{th}$ octile is

$O_i =$ Value of $\bigg(\dfrac{i(n+1)}{8}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 7$

where $n$ is the total number of observations.

Arrange the data in ascending order

62, 63, 64, 65, 66, 68, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 83

Fourth octile $O_4$

The fourth octile $O_4$ can be computed as follows:

 \begin{aligned} O_{4} &=\text{Value of }\bigg(\dfrac{4(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{4(18+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(9.5\big)^{th} \text{ observation}\\ &= \text{Value of }\big(9\big)^{th} \text{ obs.}\\ &\quad+0.5 \big(\text{Value of } \big(10\big)^{th}\text{ obs.}-\text{Value of }\big(9\big)^{th} \text{ obs.}\big)\\ &=73+0.5\big(74 -73\big)\\ &=73.5 \text{ mmHg}. \end{aligned}
Thus, $50$ % of patients had diastolic blood pressure less than or equal to $73.5$ mmHg.

Sixth octile $O_6$

The sixth octile $O_6$ can be computed as follows:

 \begin{aligned} O_{6} &=\text{Value of }\bigg(\dfrac{6(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{6(18+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(14.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(14\big)^{th} \text{ obs.}\\ &\quad+0.25 \big(\text{Value of } \big(15\big)^{th}\text{ obs.}-\text{Value of }\big(14\big)^{th} \text{ obs.}\big)\\ &=78+0.25\big(79 -78\big)\\ &=78.25 \text{ mmHg}. \end{aligned}
Thus, $75$ % of the patients had diastolic blood pressure less than or equal to $78.25$ mmHg.

## Octiles for ungrouped data Example 2

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75,89,72,78,87, 85, 73, 75, 97, 87, 84, 76,73,79,99,86,83,76,78,73.

Find the value of $O_2$ and $O_3$.

#### Solution

The formula for $i^{th}$ octile is

$O_i =$ Value of $\bigg(\dfrac{i(n+1)}{8}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 7$

where $n$ is the total number of observations.

Arrange the data in ascending order

72, 73, 73, 73, 75, 75, 76, 76, 78, 78, 79, 80, 82, 83, 84, 85, 86, 87, 97, 99

Second octile $O_2$

The second octile $O_2$ can be computed as follows:

 \begin{aligned} O_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(5.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad+0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=75+0.25\big(75 -75\big)\\ &=75 \text{ mg/dl}. \end{aligned}
Thus, $25$ % of patients had blood sugar level less than or equal to $75$ mg/dl.

Third octile $O_3$

The third octile $O_3$ can be computed as follows:

 \begin{aligned} O_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(7.875\big)^{th} \text{ observation}\\ &= \text{Value of }\big(7\big)^{th} \text{ obs.}\\ &\quad+0.875 \big(\text{Value of } \big(8\big)^{th}\text{ obs.}-\text{Value of }\big(7\big)^{th} \text{ obs.}\big)\\ &=76+0.875\big(76 -76\big)\\ &=76 \text{ mg/dl}. \end{aligned}
Thus, $37.5$ % of the patients had blood sugar level less than or equal to $76$ mg/dl.

## Octiles for ungrouped data Example 3

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137,
137, 138, 141, 143, 144, 146, 147, 152, 154, 161 

Find $5^{th}$ and $6^{th}$ octile for the above data.

#### Solution

The formula for $i^{th}$ octile is

$O_i =$ Value of $\bigg(\dfrac{i(n+1)}{8}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 7$

where $n$ is the total number of observations.

Arrange the data in ascending order

126, 129, 129, 132, 132, 133, 133, 135, 136, 137, 137, 138, 141, 143, 144, 146, 147, 152, 154, 161

Fifth octile $O_5$

The fifth octile $O_5$ can be computed as follows:

 \begin{aligned} O_{5} &=\text{Value of }\bigg(\dfrac{5(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{5(20+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(13.125\big)^{th} \text{ observation}\\ &= \text{Value of }\big(13\big)^{th} \text{ obs.}\\ &\quad+0.125 \big(\text{Value of } \big(14\big)^{th}\text{ obs.}-\text{Value of }\big(13\big)^{th} \text{ obs.}\big)\\ &=141+0.125\big(143 -141\big)\\ &=141.25 \text{ cm}. \end{aligned}
Thus, $62.5$ % of patients had height less than or equal to $141.25$ cm.

Sixth octile $O_6$

The sixth octile $O_6$ can be computed as follows:

 \begin{aligned} O_{6} &=\text{Value of }\bigg(\dfrac{6(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{6(20+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(15.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}\\ &\quad+0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=144+0.75\big(146 -144\big)\\ &=145.5 \text{ cm}. \end{aligned}
Thus, $75$ % of the patients had height less than or equal to $145.5$ cm.

## Octiles for ungrouped data Example 4

The following measurement were recorded for the drying time in hours, of a certain brand of latex paint.

3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6
3.7 2.8 4.4 4.0 5.2 3.0 4.8.

Compute $3^{rd}$ and $5^{th}$ octile for the above data.

#### Solution

Third octile $O_3$

The third octile $O_3$ can be computed as follows:

 \begin{aligned} O_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(6\big)^{th} \text{ observation}\\ &= \text{Value of }\big(6\big)^{th} \text{ obs.}\\ &=3.3 \text{ hours}. \end{aligned}
Thus, $37.5$ % of the patients had drying time less than or equal to $3.3$ hours.

Fifth octile $O_5$

The fifth octile $O_5$ can be computed as follows:

 \begin{aligned} O_{5} &=\text{Value of }\bigg(\dfrac{5(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{5(15+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(10\big)^{th} \text{ observation}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}\\ &=4 \text{ hours}. \end{aligned}
Thus, $62.5$ % of the patients had drying time less than or equal to $4$ hours.

## Octiles for ungrouped data Example 5

The rice production (in Kg) of 10 acres is given as: 1120, 1240, 1320, 1040, 1080, 1720, 1600, 1470, 1750, and 1885. Find the second octile for the given data.

#### Solution

Second octile $O_2$

The second octile $O_2$ can be computed as follows:

 \begin{aligned} O_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{8}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(10+1)}{8}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(2.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}\\ &\quad+0.75 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=1080+0.75\big(1120 -1080\big)\\ &=1110 \text{ Kg}. \end{aligned}
Thus, $25$ % of the patients had rice production less than or equal to $1110$ Kg.

## Conclusion

In this tutorial, you learned about formula for octiles for ungrouped data and how to calculate octiles for ungrouped data. You also learned about how to solve numerical problems based on octiles for ungrouped data.